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2011-02-01T00:00:01
<mdoxml version="1.0"><p>Prove that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ cannot be terms of ANY arithmetic progression.<br></br>
</p></mdoxml>
<mdoxml version="1.0"><br></br>
Thank you to <span style="font-weight: bold;">M.
Grender-Jones</span> for this solution.<br></br>
<br></br>
To show that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ cannot form part
of any arithmetic progression we give a proof by contradiction.
Suppose they can, then $$\sqrt 3- \sqrt 2 = px$$ $$\sqrt 5 - \sqrt
2 = qx$$ where $x$ is the common difference of the progression, and
$p$ and $q$ are integers. <br></br>
<br></br>
Eliminating $x$ from these two equations we get $$p(\sqrt 5 - \sqrt
2) = q(\sqrt 3 - \sqrt 2)$$ so $$q\sqrt 3 = p\sqrt 5 + (q-p)\sqrt
2.$$ We know $p$ and $q$ are integers so to simplify this
expression write $p-q = s$ where $s$ is an integer: $$q\sqrt 3 =
p\sqrt 5 + s\sqrt 2.$$ Squaring this: $$3q^2 = 5p^2 + 2s^2 +
2ps\sqrt 10.$$ Rearranging this expression gives: $$\sqrt 10 =
{3q^2 - 5p^2 - 2s^2 \over 2ps}.$$ As $\sqrt 10$ is irrational and
all the other terms in this expression are integers this is
impossible and we have reached a contradiction. Therefore our
assumption was false and $\sqrt 2$, $\sqrt 3$ and $\sqrt 5$ cannot
be terms of an AP.<br></br>
<br></br>
By the same method can you prove that $\sqrt{1}$, $\sqrt{2}$ and
$\sqrt{3}$ cannot be terms of ANY arithmetic progression? <br></br></mdoxml>
<mdoxml version="1.0"><br></br>
<span style="font-weight: bold;">Why do this problem?</span> <br></br>
It is an exercise in proof by contradiction.<br></br>
<br></br>
<span style="font-weight: bold;">Possible approach</span> <br></br>
First discuss proof by contradiction so that students appreciate
how the logic of arguments by contradiction work. You can draw on
this article on <a href="http://nrich.maths.org/public/viewer.php?obj_id=4717&part=index">
Proof by Contradiction.</a> <br></br>
<br></br>
<div>Then discuss the proof that $\sqrt 2$ is irrational.</div>
<br></br>
<div>The students can work with the interactivity<a href="http://nrich.maths.org/public/viewer.php?obj_id=1404&part=index">
Proof Sorter</a> and perhaps some of them might read <a href="http://nrich.maths.org/public/viewer.php?obj_id=4717&part=index">
this article</a> which was written by two undergraduates.</div>
<br></br>
<div><span style="font-weight: bold;">Key Question</span></div>
<br></br>
<div>If the difference between$\sqrt 2$ and $\sqrt 3$ is an integer
multiple of the common difference in an arithmetic series, and the
difference between $\sqrt 3$ and $\sqrt 5$ is also an integer
multiple of that common difference, can you use these two facts to
write down two expressions, eliminate the unknown common difference
and then find an impossible relationship?</div>
<br></br>
<div><span style="font-weight: bold;">Possible support</span></div>
<div><a href="http://nrich.maths.org/public/viewer.php?obj_id=1404&part=index">
Proof Sorter</a> and <a href="http://nrich.maths.org/public/viewer.php?obj_id=4717&part=index">
article</a></div>
<br></br>
<div><span style="font-weight: bold;">Possible
extension</span></div>
<br></br>
<div>Can you prove that $\sqrt{1}$, $\sqrt{2}$ and $\sqrt{3}$
cannot be terms of ANY arithmetic progression?</div>
<br></br>
<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
Try a proof by contradiction. Suppose that the three irrational
numbers do occur in some arithmetic series. Can you then go on to
reach a contradiction?<br></br>
<br></br>
<div>It helps to have seen a proof that $\sqrt 2$ is irrational and
to appreciate how the logic of arguments by contradiction
work.</div>
<br></br>
<div>See <a href="http://nrich.maths.org/public/viewer.php?obj_id=1404&part=index">
Proof Sorter</a> and, for some further reading on proofs by
contradiction, see <a href="http://nrich.maths.org/public/viewer.php?obj_id=4717&part=index">
this article</a> written by two undergraduates.</div>
<br></br>
<div>Then you only need to know the definition of an arithmetic
series to do this problem. If the difference between $\sqrt 2$ and
$\sqrt 3$ is an integer multiple of the common difference in an
arithmetic series, and the difference between $\sqrt 3$ and $\sqrt
5$ is also an integer multiple of that common difference, can you
use these two facts to write down two expressions, eliminate the
unknown common difference and then find an impossible relationship?
</div>
<br></br>
<br></br>
<br></br>
<br></br></mdoxml>
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Be reasonable
Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic
progression.
Sequences, Functions and Graphs
Arithmetic sequence
Using, Applying and Reasoning about Mathematics
Proof by contradiction
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